Ascending Pea Pattern generator

Question

For those who are not familiarized with the Pea Pattern, it is a simple mathematical pattern.

There are multiple variations of this pattern, but we will focus in one:

Ascending Pea Pattern

It looks like this:

1
11
21
1112
3112
211213
...

It seems really hard to get the following line, but it is really easy. The way to get the next line is by counting the number of times a digit have repeated on the previous line (start counting with the lowest, to largest):

one
one one
two ones
one one, one two
three ones, one two
two ones, one two, one three
...

Requirements/Rules:

• We will start at 1
• It will be a snippet
• There must be a way to specify the number of lines generates (e.g 5 will give the first 5 lines)
• The code should be as short as possible
• It must start counting from lowest to largest (the Ascending variation)

Answer 1

APL, 32 characters

⍪⌽({⍵,⍨,/{⌽⍵,+/⍵⍷d}¨⍳⌈/d←⊃⍵}⍣⎕)1

This generates lines starting from 0 (i.e. 0 generates 1, 1 generates 1 followed by 1 1, etc.), as specified by user input. I used Dyalog APL for this, and ⎕IO should be set to its default of 1.

Example:

      ⍪⌽({⍵,⍨,/{⌽⍵,+/⍵⍷d}¨⍳⌈/d←⊃⍵}⍣⎕)1
⎕:
      0
1

      ⍪⌽({⍵,⍨,/{⌽⍵,+/⍵⍷d}¨⍳⌈/d←⊃⍵}⍣⎕)1
⎕:
      13
               1 
             1 1 
             2 1 
         1 1 1 2 
         3 1 1 2 
     2 1 1 2 1 3 
     3 1 2 2 1 3 
     2 1 2 2 2 3 
     1 1 4 2 1 3 
 3 1 1 2 1 3 1 4 
 4 1 1 2 2 3 1 4 
 3 1 2 2 1 3 2 4 
 2 1 3 2 2 3 1 4 
 2 1 3 2 2 3 1 4

Once I get some more time, I'll write up an explanation. ⍨

Answer 2

Python (2.x), 81 80 characters

l='1'
exec"print l;l=''.join(`l.count(k)`+k for k in sorted(set(l)))\n"*input()

All tips or comments welcome!

usage: python peapattern.py
15 # enter the number of iterations
1
11
21
1112
3112
211213
312213
212223
114213
31121314
41122314
31221324
21322314
21322314
21322314

Answer 3

J, 60 46 39 26 characters

1([:,(#,{.)/.~@/:~)@[&0~i.

Edit 3: Came up with a much nicer way of expressing this.

1([:;[:|."1[:/:~~.,.[:+/"1[:~.=)@[&0~i.

Edit 2: Finally found a way to move the argument to the end of the sequence and get rid of the unnecessary assignment stuff.

Previously:

p=.3 :'([:,[:|."1[:/:~~.,.[:+/"1[:~.=)^:(i.y)1

Edit 1: Fixes the output which should be y rows rather than the yth row. Also shortens things a bit. Shame about the 0s, can't seem to get rid of the damn things.

Usage:

   1([:,(#,{.)/.~@/:~)@[&0~i. 1
1

   1([:,(#,{.)/.~@/:~)@[&0~i. 6
1 0 0 0 0 0
1 1 0 0 0 0
2 1 0 0 0 0
1 1 1 2 0 0
3 1 1 2 0 0
2 1 1 2 1 3

   1([:,(#,{.)/.~@/:~)@[&0~i. 10
1 0 0 0 0 0 0 0
1 1 0 0 0 0 0 0
2 1 0 0 0 0 0 0
1 1 1 2 0 0 0 0
3 1 1 2 0 0 0 0
2 1 1 2 1 3 0 0
3 1 2 2 1 3 0 0
2 1 2 2 2 3 0 0
1 1 4 2 1 3 0 0
3 1 1 2 1 3 1 4

Admittedly the usage is uglier now, but prettiness isn't the name of the game here...

Answer 4

Haskell, 116

import Data.List
main=interact(unlines.map(show=<<).($iterate((>>=
 \x->[length x,head x]).group.sort)[1]).take.read)

Usage:

$ runhaskell pea.hs <<< 15
1
11
21
1112
3112
211213
312213
212223
114213
31121314
41122314
31221324
21322314
21322314
21322314

Answer 5

Common Lisp, 140 characters

(defun m(x)
  (labels((p(l n)
    (if(= 0 n)
       nil
       (cons l
             (p(loop for d in(sort(remove-duplicates l)#'<)
                  append(list(count d l)d))
               (1- n))))))
    (p'(1) x)))

This is Lisp, so the function returns a list of lists. (m x) generates X sublists.

Answer 6

Perl, 83

I'm pretty sure some Perl guru could outdo this, but here goes:

$_++;$n=<>;for(;$n--;){print$_.$/;$r='';$r.=length($&).$1 while(s/(.)\1*//);$_=$r;}

Expanded:

$_++;$n=<>;

for(;$n--;)
{
    print $_.$/;

    $r='';$r .= length($&).$1 while (s/(.)\1*//);  # The magic
    $_=$r;
}

Number of rows is passed in via STDIN.

Answer 7

Mathematica, 70

NestList[FromDigits@TakeWhile[DigitCount@#~Riffle~Range@9,#>0&]&,1,#]&